Community member
https://youtu.be/OQCjakjCJu4?si=wqTQV6d1yjyb_0E4
In this one shoot at 4.18.00 I think that example is not LALR because in I4 you took wrong production A->d,c but it is actually A->d,a so their collision occurs
KG AI MentorKG AI Mentor
Quick AnswerYour observation about the production choice is correct—using A→d,a instead of A→d,c would indeed cause a shift-reduce conflict in that state. Key StepsIn LALR(1), lookahead sets are merged from LR(0) states, which can introduce conflicts if lookaheads differ. The production A→d,a and A→d,c must be checked against the specific lookahead symbols in state I4. A collision occurs when both a reduce action and a shift action are required for the same symbol. — KG AI Mentor 🤖
@kg_ai_mentor when you fetch A->.d,a in I4 you have to write A->d.,a but you written A->d.,c in I4 so it is wrong
KG AI MentorKG AI Mentor
Quick AnswerThe state I3 (labeled I4 in the image) correctly uses FOLLOW(A)={c,a}, so there is no collision. Key StepsIn state I3, the item is $S \to bA\cdot c$, which requires FOLLOW(A) for reduction. The follow set of A is derived from the grammar: $S \to Aa$ gives 'a', and $S \to bAc$ gives 'c'. Therefore, both reductions A → d on input 'a' and 'c' are valid and distinct. Since the lookahead sets {a} and {c} do not overlap, this is indeed a valid LALR(1) construction. — KG AI Mentor 🤖